Chemistry, asked by Shipri3563, 1 year ago

Write tge mechanism for the prepration of ethonal from ethane .

Answers

Answered by wikremsinghp4v470
0

he dehydration of ethanol:-

Dehydration of propan-2-ol, you will know that it involves the formation of a carbocation (a carbonium ion).

If ethanol used the same mechanism, you would get a primary carbocation formed, CH3CH2 , but this is much less stable than a secondary or tertiary carbocation.

That would lead to a very high activation energy for the reaction.

The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy.

We are going to discuss the mechanism using sulphuric acid.

In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. That is exactly the same as happens with propan-2-ol and the other secondary and tertiary alcohols.

In the mechanism we have already looked at with propan-2-ol, the next thing to happen was loss of water to form a carbocation, followed by removal of a hydrogen ion from the carbocation and the formation of a double bond.

In this case, instead of happening in two separate steps, this all happens at the same time in one smooth operation. By doing that, you avoid the formation of an unstable primary carbocation.


Answered by sunilsihag403
0

he dehydration of ethanol:-


Dehydration of propan-2-ol, you will know that it involves the formation of a carbocation (a carbonium ion).


If ethanol used the same mechanism, you would get a primary carbocation formed, CH3CH2 , but this is much less stable than a secondary or tertiary carbocation.


That would lead to a very high activation energy for the reaction.


The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy.


We are going to discuss the mechanism using sulphuric acid.


In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. That is exactly the same as happens with propan-2-ol and the other secondary and tertiary alcohols.


In the mechanism we have already looked at with propan-2-ol, the next thing to happen was loss of water to form a carbocation, followed by removal of a hydrogen ion from the carbocation and the formation of a double bond.


In this case, instead of happening in two separate steps, this all happens at the same time in one smooth operation. By doing that, you avoid the formation of an unstable primary carbocation

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